# Does direction of current in Node Voltage method circuit analysis matter?

I am trying to solve the following circuit using the node voltage method, but I'm having issues with figuring out how current is supposed to flow in and out of nodes. I understand that the current arrows shown are just a reference for the direction in which you should write the answer (i.e., if $i_a$ is a positive current going right to left, the answer will be negative). When I try to write my equations, I set the bottom wire as ground and get the following by going through the steps of node voltage method:

\begin{align*} \frac{v_1-v_3}{5}-\frac{v_3}{60}-\frac{v_4-v_3}{4} &= 0,\\ \frac{v_2-v_4}{10}-\frac{v_4}{80}+\frac{v_3-v_4}{4} &= 0. \end{align*}

(Here $v_3$ is the voltage at the top left dot, $v_4$ at the top right dot.)

As I understand it, the sign on unknown currents is arbitrary since if the sign is wrong, then the result will simply be negative. However, in the way my signs are set, I get the wrong currents despite my equations matching the solution in all other regards as far as I can tell.

Is there an issue with my equations that I can’t see, or am I misunderstanding something about how currents work?

The values I used for $v_1$ and $v_2$ were $112\,\mathrm V$ and $330\,\mathrm V$ respectively.

One issue here is the inconsistency with a current direction.

For instance, if you decided to consider currents flowing in as positive (as I believe you did), it is better to show the terms associated with these currents as positive everywhere. In your case, current Ie is flowing out, but is presented as positive.

Another issue is the inconsistency with deriving a current from node voltages.

If you assume that a current flows from a to b, it has to be expressed as (Va-Vb)/R. In your case, you assumed that Ic flows from node 3 to node 4, but you've expressed it as (V4-V3)/R.

NOTE: if you feel like doing the opposite, i.e., expressing it as (Vb-Va)/R, you should do to it the same way everywhere, which would essentially mean that all your arrows have to be flipped.

The term (v2-v4)/10 has two inconsistencies in it and, as a result, it is technically correct, but it would be easier to understand the equation if it was written as -(v4-v2)/10.

• The formulas I wrote don't exactly follow the direction arrows shown on the diagram (which may have been my first mistake, admittedly) and I decided to write Ie as a current going in. Would this affect the result somehow? Other than making the sign for Ie negative – Dan Apr 1 '18 at 0:18
• @Dan It should not affect the magnitude of Ie , but when the equations don't exactly match the diagram, it is easier to make a mistake. – V.F. Apr 1 '18 at 0:30

I would write the first equation as the sum of the currents entering (or leaving) the junction equal to zero.
As current flows from high potential to low potential the currents will all be of the following form as they are entering the node at potential $v_3$ which is at the “lower” potential.

$$\frac{v_{\rm x}-v_3}{R}$$ This gives

$$\frac{v_1-v_3}{5}+\frac{0-v_3}{60}+\frac{v_4-v_3}{4} = 0$$

Which is not quite the same as your first equation $$\frac{v_1-v_3}{5}-\frac{v_3}{60}-\frac{v_4-v_3}{4} = 0$$

I have assumed that the bottom node is at a potential of zero and completely ignored any current direction arrows that you have drawn in your diagram.

You can do the same for the other node at potential $v_4$ and then solve for $v_3$ and $v_4$.

Then, for example, $i_{\rm b} = \dfrac{v_3-0}{60}$ as the direction of the current as shown in the diagram is from the node at potential $v_3$ to the node at potential zero.

• So for my second equation, would I then use that same method but use V4 instead of V3? – Dan Apr 1 '18 at 0:21
• @Dan Yes, the currents into that node will be of the form $\frac{v_{\rm x}-v_4}{R}$ and I think that you should find $i_{\rm e} = - \frac{213}{16}\,\rm A$ ie actually flowing in the opposite direction to that shown in your diagram. – Farcher Apr 1 '18 at 6:44